Q2. If mass of the moving body ismuch greater than the mass of the body at rest than what is the approximatevilocity of the moving body after head-on collision?

Ans2. When the moving mass is muchgreater than the mass at rest then after the collision the heavier mass keepson moving with the same velocity and in the same direction.

Q3. At what point the potentialenergy of a body is taken to be zero?

Ans3. The potential energy of the bodyat the surface of the earth is taken to be zero (Potential energy = mgh, whereh is the height of the body from the surface.).

Q4. Does the work done on a body by aforce depend upon the path followed by it?

Ans4. May or may not be. If the force isconservative then it does not depend but if it is non-conservative (friction)then it depends.

Q5. If a body hits the ground from aheight h1 and rebounds to a height h2 after havinginelastic collision with the ground then what is the coefficient ofrestitution?

Ans5. e = Ö(h2/h1)

Q6. A body hits the ground with 50m/s velocity and has inelastic collision with the ground then with whatvelocity it will rebound if the coefficient of restitution is 0.2.

Ans6. Coefficient of restitution, e = v2/v1where v1 is the velocity with which the body hits the ground and v2is the velocity of rebound.
e = 0.2 = v2/50, so v2 = 10 m/sec.

Q7. A body at rest explodes in threefragments. Is it possible that two equal parts move in mutually perpendiculardirections with the same velocity and third mass moves midway between the two?

Ans7. No, it is not possible becausemomentum before the explosion is zero and after it also the momentum must bethe same. In the given situation the third particle must go in a directionopposite to the resultant of first two parts.

Q8. If the speed of a moving vehicleis increased by 200% then how much should be the change in the retarding forceto stop the vehicle over half the previous distance?

Ans8. (1/2)mv2 = F.S where Fis the retarding force and S is the distance over which the vehicle comes torest. When 'v' is increased by 200% then K.E. increases by 800%. As S is halvedthen F should be made 16 times.

Q9. If 20 Joules of work is done in compressinga spring from 0 cm to 6 cm then find the work done in compressing the same from3cm to 6 cm.

Q1. A ball is dropped from rest at aheight of 20m. If it loses 30% of its kinetic energy on striking the ground,what is the height to which it bounces? How do you account for this loss inkinetic energy?

Ans1. Suppose the ball acquires avelocity 'v' after falling through a height of 20m.
Because the ball is dropped from rest, hence u = 0.
Hence, v2 = u2 + 2as
= 0 + (2 x 10 x 20) = 400
So, v = 20 m/s
Kinetic energy of the ball just before hiting the ground
= (1/2)mv2 = (1/2)m(400) = 200m Joule
Because the ball loses 30% of the kinetic energy on striking the ground, hencekinetic energy retained by the ball after striking the ground = 70% of 200m J
= 140m J
The energy loss is due to the inelastic collision with the ground.

Q2. Why is no energy being consumedin planetary motion.

Ans2. A planet is a heavenly body whichrevolves round the star (the sun). The force which is responsible for circularmotion, is called centripetal force. The direction of the centripetal force isalways towards the centre. Thus, the angle between force F and displacement Sis q = 90o at every point.Work done in moving planet, W = F.S = FSCosq = FS Cos90o.So, W = 0. Hence,no energyis being consumed in planetary motion.

Q3. How will you justify thathydro-electric power plant is an illustration of law of conservation of energy?

Ans3. A hydroelectric power-plant isused in generating electric energy (Power).The potential energy of water storedat a height is converted into K.E. when water is made to rush down. This fallof water is used to rotate the turbine and the coil and armature of generatoris rotated and electricity is produced . Thus, the K.E. of the fall of water isconverted into electrical form of energy. Hence the hydroelectric power-plantis an example of law of conservation of energy.

Q4. If a body is dropped from aheight of 40m then after 3 inelastic collisions with the ground to which heightthe body will rise? (given: Coefficient of restitution = 0.5)

Ans4. If the body is dropped from aheight of 'H' and 'e' is the coefficient of restitution then after 'n'inelastic collisions with the ground the body rises to a height 'h' given by
h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m

Q5. A truck and a car are moving withthe same K.E. on a straight line road. If their engines are made off at thesame time, which one of them will stop at a lesser distance.

Ans5. Given : (1/2)m1v12= (1/2)m2v22 .............(i)
If force of brakes be the same then m1a1 = m2a2..........(ii)
If truck stops over a distance S1 then v12 =2a1S1 ........(iii)
If car stops over a distance S2 then v22 = 2a2S2.........(iv)
From (i) and (ii)
(1/2)m1v12 = (1/2)m2v22...........(v)
From (ii) and (v)
v12/a1 = v22/a2............(vi)
From (iii), (iv) and (vi)
2a1S1/a1 = 2a2S2/a2.
S1 = S2
Hence distances covered S1 and S2 are equal.

Q6. The power of a pump motor is 4KW. How much water in kg/minute can it raise a height of 20m? (g = 10 m/s2)

Q7. A rod of length 3m is suspendedvertically from a fixed point. It is given an angular displacement of 60oin the vertical plane. If its mass per unit length is 2 kg then find the workdone?

Ans7. Let 'm' be the mass of the rod and'l' be its length then
m = 2 x 3 = 6 kg
If the rod is displaced through an angle qthen the work done on it, W = mg(l/2)(1 - Cosq).
The effective length of the rod is taken to be (l/2) because in uniformdistribution of mass the centre of mass is at the geometric centre so
W = 6 x 10 x (3/2)(1 - Cos60) =45 J.

Three mark questions with answers

Q1. When a mass m2is at rest and mass m1 moving with velocity u1 hits itelastically, show that the fraction of the momentum transferred to the mass atrest is 2n/(1 + n) where n is ratio of the masses.

Q2. Is it possible for work to bepositive negative or zero? "Explain with example.

Ans2. Work is a scalar quantity which isgiven by the scalar product of the force applied F and the displacement S movedby body i.e.
W = F.S,
W = FS Cos q ..........(1).
If q = 0. The work is maximum.
It remains +ve for the angle q between 0oto 90oor, q lying between 270o and360oi.e., if the displacement is in a directionoppisite to which the force is applied.
Thus work is +ve if Cos q is +ve. The work done will be -veif Cosq is -ve i.e.q lies between 90o to 270o.
If q = 90o then Cos 90o= 0.
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero

Q3. The bob of a simple pendulum isreleased from a horizontal position. If the length of the pendulum is 2m, whatis the speed with which the bob arrives at the lowermost point? Given that itdissipates 10% of its initial energy against air resistance?

Q6. When water is flowing through apipe then its velocity changes by 5%, find the change in the power of water?

Ans6. Power = Force xVelocity = Rate of change of momentum x velocity ={(mass/time) x velocity} xvelocity = {(adv) x v} x v =adv3 where 'a' is area of cross section, 'd' is the density of waterand 'v' is the velocity of flow of water.
Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv3 (k is a constant and is equal to 'ad'.)Taking log on both sides
log P = 3log v + log k
Differentiating on both sidesDP/P = 3.Dv/v
percentage change in power, DP/P x100 = 3 x 5%
= 15%.

Q7. The kinetic energy of rushing outwater from a dam is used in rotating a turbine. The pipe through which water isrushing is 2.4 meters and its speed is 12 m/sec. Assuming that whole of kineticenergy of the water is used in rotating the turbine, calculate the currentproduced if efficiency of the dynamo is 60% and the station transmits power at240 kV. Density of water = 103 kg/m3.

Q1. A vehicle of mass m isaccelerated from rest when a constant power P is supplied by its engine; showthat :
(a) The velocity is given as a function of time by
v = (2Pt/m)1/2
(b) The position is given as a function of time by
s = (8P/9m)1/2t3/2.
(c) What is the shape of the graph between velocity and mass of the vehicle ifother factors remain same?
(d) What is the shape of the graph between displacement and power?

Q2. A simple pendulum of mass m and lengthl swings back and forth up to a maximum angle q0 with the vertical. When at an angle q, what is its (a) potential energy, (b) kinetic energy, (c) speed, and(d) tension?

Q3.What do youmean by work in the language of physics? Give its absolute and gravitationalunits. Give two illustrations of zero work, negative work and positive work.

Ans.(Try yourself).

Q4.How will youfind work done by a variable force mathematically and graphically?

Ans.(Try yourself).

Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.

Ans.(Try yourself).

Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.

Ans.(Try yourself).

Q7.If a body iskept on the top of a rough inclined plane, find the expression for
(i) work done in bringing it down to the bottom of the plane with constantvelocity
(ii) work done in moving it up the plane with constant acceleration
(iii) work done in moving it down the plane with constant acceleration.